Question

The parabola $x^2=4ay$ makes an intercept of length $\sqrt{40}$ units on the line $y=1+2x$ then a value of $4a$ is

• A. 2
• B. -2
• C. -1
• D. 4

Answer 1

Answer: (B)

Since length of the chord intercepted from the line
$y=mx+c$ by the parabola $x^2=4ay$ is
$4\sqrt{a\left(1+m^2\right)\left(c+a{m}^2\right)}$
So, $\sqrt{40}=4\sqrt{a(1+4)(1+a(4))}$
$40=16(5a(1+4a))$
$\Rightarrow 1=2a(1+4a)$
$\Rightarrow 8a^2+2a-1=0$
$\Rightarrow 8a^2+4a-2a-1=0$
$\Rightarrow 4a(2a+1)-1(2a+1)=0$
$\Rightarrow a=\frac{1}{4}$ or $-\frac{1}{2}$
$\Rightarrow 4a=1$ or$-2$